PV Diagrams and Ideal Gasses -- I ONLY NEED help answering PART D! A graph of Pr

Question

PV Diagrams and Ideal Gasses -- I ONLY NEED help answering PART D!

A graph of Pressure versus Volume is given for an ideal gas within a closed container. The temperature of the system at point A is 20°C. (Figure 1) Part A What is the temperature at point B? 73.3 K Submit Correct Part B What is the temperature at point C? 366 K Correct Part C How many molecules of gas are inside the container? 1.00×10^26 molecules Correct

ONLY NEED PART D CALCULATED

Part D What is the total work done on the system to go from point A, to B, to C, and back to A? I took area of the rectangle under B to C (done by a gas) and subtracted it from area under triangle (done to the gas) for -202,600 J (incorrect) I also tried without the negative.

Explanation / Answer

This is a thermodynamic problem, so we use the equation work termodinámica

W = - ViVfP dV

We calculate the work in every part of the path

A ---B

Vi = Vf = 1

Vf-Vi =0 dV= 0

Wab = 0

b — > C

P= cte = 1 Atm

W = - ViVf P dV W = - P ViVf dV

W = - P ( Vf -Vi)

W = - 1 ( 5-1)

W = -4 Atm m³

1 atm = 1,013 105N/m²

Wbc = -4.052 105J

C — > A

Note that for this work a linear relationship in the graph is proposed, which use the equation to work must first find this line

P-Po = m (V-Vo)

m It is the slope

m = (P2-P1)/(V2-V1) point 2 (4,1) point 1 (1,5)

m =( 4-1)/(1-5)

m = 3/(-4) = 0.75 Atm/ m³

m = -0.75

(P-1)= -0.75 (V-5)

P = 1- 0.75V + 3.75

P = 4.75 – 0.75 V

W = - ViVf P dV

W = - ViVf(4.75 -0.75 V) dV

W = -4.75 ViVfdV + 0.75 ViVfV dV

W = -4.75 V ++ 0.75 ½ V2

We must be very careful evaluation of the integral as identifying which is the final value depends on the direction of travel, if we look at the diagram,

final point ( 4 atm, 1 m³)

starting point (1 atm, 5 m³)

substituting

W = -4.75 (1-5) + 0,375 (12– 52) W = 8.75 – 9

W= - 0,25 atm m³ w= -0.25 1,013 105N/m² m³

W = - 0,25325 105J

Now we must add the different jobs

Wt = Wab +Wbc +Wca

Wt= 0 + (-4.052 105) + (-0.25325 105)

Wt = - 4.31 105J

Be very careful with the signs of each job

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